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Chapter 7: Problem 8

In a random sample of 80 components of a certain type, 12 are found to be defective. a. Give a point estimate of the proportion of all such components that are not defective. b. A system is to be constructed by randomly selecting two of these components and connecting them in series, as shown here. The series connection implies that the system will function if and only if neither component is defective (i.e., both components work properly). Estimate the proportion of all such systems that work properly. [Hint: If \(p\) denotes the probability that a component works properly, how can \(P\) (system works) be expressed in terms of \(p\) ?] c. Let \(\hat{p}\) be the sample proportion of successes. Is \(\hat{p}^{2}\) an unbiased estimator for \(p^{2} ?\)

Short Answer

Expert verified
a. 0.85. b. 0.7225. c. No, \( \hat{p}^2 \) is not unbiased for \( p^2 \).

Step by step solution

01

Calculate Point Estimate for Non-Defective Proportion

First, we find the number of non-defective components in the sample. There are 80 components total, and 12 are defective. The number of non-defective components is 80 - 12 = 68. The sample size is 80, so the point estimate for the proportion of non-defective components is given by dividing the non-defective count by the total count: \( \hat{p} = \frac{68}{80} = 0.85 \).
02

Estimate System Working Proportion

In a two-component series system, both components must work for the system to function. If \( p \) is the probability that a component is non-defective, then the probability that both work is \( p^2 \). Based on our point estimate from Step 1, \( \hat{p} = 0.85 \). Thus, the estimated proportion of functioning systems is \( \hat{p}^2 = (0.85)^2 = 0.7225 \).
03

Evaluate Unbiased Estimator Property

An estimator is unbiased if its expected value equals the parameter it is estimating. Here, we want to determine if \( \hat{p}^2 \) is an unbiased estimator for \( p^2 \). The bias is calculated using: \( E(\hat{p}^2) - p^2 \). For large samples, \( \hat{p}^2 \) is not an unbiased estimator of \( p^2 \) due to its nonlinear nature, implying \( \hat{p}^2 \) can be biased, particularly for small sample sizes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Proportion
In statistics, the sample proportion is a fundamental concept that serves as an estimator for the true proportion within a population. When you take a random sample, you aim to gauge the proportion of a certain characteristic in that sample, which can provide insight into the overall population.
A sample proportion is represented by \( \hat{p} \). In our exercise, we started by calculating the sample proportion of non-defective components. We found that out of 80 components, 68 were non-defective.
Thus, the sample proportion \( \hat{p} \) was calculated as \( \frac{68}{80} = 0.85 \). This 0.85 figure provides an estimate of the proportion of all components that are not defective.
  • The calculation involves counting the number of successes (non-defective components) and dividing by the total sample size.
  • It is critical for providing initial estimates that are used for further probability and statistical analysis.
Unbiased Estimator
An unbiased estimator is crucial in statistical analysis as it ensures that estimates reflect the true population parameters without systematic error. In essence, an unbiased estimator's expected value equals the parameter it estimates.
For the exercise, the question arises whether \( \hat{p}^2 \), which estimates the probability that both components are non-defective in a series system, is unbiased for \( p^2 \).
While \( \hat{p} \) is an unbiased estimator for \( p \) itself, complications arise when squaring this estimator, because the expected value doesn't conveniently equal \( p^2 \).
  • Because \( \hat{p}^2 \) is non-linear, its unbiasedness depends heavily on sample size. Larger samples reduce the bias.
  • It highlights the importance of understanding estimator properties when interpreting results.
In conclusion, \( \hat{p}^2 \) isn't unbiased for \( p^2 \), especially in small samples.
Probability
Probability provides a way to foresee how likely an event is to occur. It is foundational to understanding and analyzing risk and uncertainty in any system.
In our exercise, probability is essential to estimate the likelihood of a system working properly.
Given \( p \) as the probability that an individual component is non-defective, the probability that the series system is functioning (i.e., both components work) is \( p^2 \). This is because each component must independently work for the entire system to function.
  • The calculation assumes independence, where the outcome of one component does not affect the other.
  • Probabilities are often represented as values between 0 and 1, where 1 indicates certainty of occurrence.
This framework helps us assess how reliable systems are based on individual component probabilities.
Series System Reliability
Understanding series system reliability is key when dealing with systems where multiple components must work for the whole system to function. In a series connection, like the one considered in our exercise, if any one component fails, the entire system fails.
Reliability in this context directly relates to the probability that all components in the system are functioning properly. If each component has a probability \( p \) of working correctly, the reliability of the two-component series system can be represented as \( p^2 \).
  • This reliability metric is crucial when designing systems to ensure high performance standards.
  • It's important to consider that each additional component added to the series increases the complexity and decreases the overall reliability, unless steps are taken to ensure higher individual component reliability.
Thus, achieving a reliable system necessitates not only knowledge of individual component reliability but also careful consideration of system design.

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Most popular questions from this chapter

Suppose a certain type of fertilizer has an expected yield per acre of \(\mu_{1}\) with variance \(\sigma^{2}\), whereas the expected yield for a second type of fertilizer is \(\mu_{2}\) with the same variance \(\sigma^{2}\). Let \(S_{1}^{2}\) and \(S_{2}^{2}\) denote the sample variances of yields based on sample sizes \(n_{1}\) and \(n_{2}\), respectively, of the two fertilizers. Show that the pooled (combined) estimator $$ \hat{\sigma}^{2}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2}}{n_{1}+n_{2}-2} $$ is an unbiased estimator of \(\sigma^{2}\).

The principle of unbiasedness (prefer an unbiased estimator to any other) has been criticized on the grounds that in some situations the only unbiased estimator is patently ridiculous. Here is one such example. Suppose that the number of major defects \(X\) on a randomly selected vehicle has a Poisson distribution with parameter \(\lambda\). You are going to purchase two such vehicles and wish to estimate \(\theta=P\left(X_{1}=0, \quad X_{2}=0\right)=e^{-2 \lambda}\), the probability that neither of these vehicles has any major defects. Your estimate is based on observing the value of \(X\) for a single vehicle. Denote this estimator by \(\hat{\theta}=\delta(X)\). Write the equation implied by the condition of unbiasedness, \(E[\delta(X)]=e^{-2 \lambda}\), cancel \(e^{-\lambda}\) from both sides, then expand what remains on the right-hand side in an infinite series,and compare the two sides to determine \(\delta(X)\). If \(X=200\), what is the estimate? Does this seem reasonable? What is the estimate if \(X=199 ?\) Is this reasonable?

Let \(p\) denote the proportion of all individuals who are allergic to a particular medication. An investigator tests individual after individual to obtain a group of \(r\) individuals who have the allergy. Let \(X_{i}=1\) if the \(i\) th individual tested has the allergy and \(X_{i}=0\) otherwise ( \(i=1,2,3, \ldots\) ). Recall that in this situation, \(X=\) the number of nonallergic individuals tested prior to obtaining the desired group has a negative binomial distribution. Use the definition of sufficiency to show that \(X\) is a sufficient statistic for \(p\).

Each of 150 newly manufactured items is examined and the number of scratches per item is recorded (the items are supposed to be free of scratches), yielding the following data: \begin{tabular}{llllllllll} Number of scratches per item & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline Observed frequency & 18 & 37 & 42 & 30 & 13 & 7 & 2 & 1 \\ \hline \end{tabular} Let \(X=\) the number of scratches on a randomly chosen item, and assume that \(X\) has a Poisson distribution with parameter \(\lambda\). a. Find an unbiased estimator of \(\lambda\) and compute the estimate for the data. [Hint: \(E(X)=\lambda\) for \(X\) Poisson, so \(E(\bar{X}=\) ?)] b. What is the standard deviation (standard error) of your estimator? Compute the estimated standard error. [Hint: \(\sigma_{X}^{2}=\lambda\) for \(X\) Poisson.]

Here is a result that allows for easy identification of a minimal sufficient statistic: Suppose there is a function \(t\left(x_{1}, \ldots, x_{n}\right)\) such that for any two sets of observations \(x_{1}, \ldots, x_{n}\) and \(y_{1}, \ldots, y_{n}\), the likelihood ratio \(f\left(x_{1}, \ldots, x_{n} ; \theta\right) / f\left(y_{1}, \ldots, y_{n} ; \theta\right)\) doesn't depend on \(\theta\) if and only if \(t\left(x_{1}, \ldots, x_{n}\right)\) \(=t\left(y_{1}, \ldots, y_{n}\right)\). Then \(T=t\left(X_{1}, \ldots, X_{n}\right)\) is a minimal sufficient statistic. The result is also valid if \(\theta\) is replaced by \(\theta_{1}, \ldots, \theta_{k}\), in which case there will typically be several jointly minimal sufficient statistics. For example, if the underlying pdf is exponential with parameter \(\lambda\), then the likelihood ratio is \(\lambda^{\sum x_{i}-\Sigma y_{i}}\), which will not depend on \(\lambda\) if and only if \(\sum x_{i}=\sum y_{i}\), so \(T=\sum x_{i}\) is a minimal sufficient statistic for \(\lambda\) (and so is the sample mean). a. Identify a minimal sufficient statistic when the \(X_{i}\) 's are a random sample from a Poisson distribution. b. Identify a minimal sufficient statistic or jointly minimal sufficient statistics when the \(X_{i}\) 's are a random sample from a normal distribution with mean \(\theta\) and variance \(\theta\). c. Identify a minimal sufficient statistic or jointly minimal sufficient statistics when the \(X_{i}\) 's are a random sample from a normal distribution with mean \(\theta\) and standard deviation \(\theta\).
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